The addition, removal, or substitution of nitrogenous bases in a DNA molecule may be caused by . Synapsis C. Disjunction B. Hybridization D. Mutation

Answer:

The answer is "2.459".

Explanation:

The cell potential or emf norm  = oxidation pot. of anode -oxidation pot. of cathode

When the pot oxidizer is the opposite of a red pot size it's been transformed to either the redox reaction so that anode was its higher oxide pot material that means (AL) so,

[tex]\to emf=1.66-(-0.799) \\\\ \to emf=1.66+0.799) \\\\ \to emf =2.459[/tex]

Answer:

12 M

Explanation:

The reaction between HNO₃ and KOH is:

First we calculate how many KOH moles reacted with the diluted HNO₃ sample, using the given volume and concentration:

As 1 KOH mol reacts with 1 HNO₃ mol, in 25 mL of the diluted HNO₃ solution there are 30 HNO₃ mmoles.

With that information in mind we can calculate the HNO₃ concentration in the diluted solution:

Finally we can use the C₁V₁=C₂V₂ formula to calculate the concentration of the original solution:

Answer:

Nitrogen cycle

Explanation:

Answer:

[tex]T_F=17.56\°C[/tex]

Explanation:

Hello there!

In this case, for this calorimetry problem, it is possible for us to realize that the heat lost by the hot silver is gained by the cold far whose specific heat is 3.94 J/g°c, so we can write:

[tex]-Q_{Ag}=Q_{fat}[/tex]

Which can be written in terms of mass, specific heat and temperature as shown below:

[tex]-m_{Ag}C_{Ag}(T_F-T_{Ag})=m_{fat}C_{fat}(T_F-T_{fat})[/tex]

In such a way, solving for the final temperature, we obtain:

[tex]T_F=\frac{m_{Ag}C_{Ag}T_{Ag}+m_{fat}C_{fat}T_{fat}}{m_{Ag}C_{Ag}+m_{fat}C_{fat}}}[/tex]

Then, we plug in the given data to obtain:

[tex]T_F=\frac{14.16g*0.2165J/g\°C*133.5\°C+250g*3.94J/g\°C*17.20\°C}{14.16g*0.2165J/g\°C+250g*3.94J/g\°C} \\\\T_F=17.56\°C[/tex]

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Answer:

65.1 grams

Explanation:

The percentage of impurities in the contaminated CaCl₂ sample is:

Meaning that the purity of the sample is (100 - 7) 93%.

Now we calculate the mass of pure CaCl₂ in the contaminated sample:

Thus the answer is 65.1 grams, as that amount is how much CaCl₂ is in the sample and it would be impossible to obtain more than that.

Answer:

1.34 mol

Explanation:

Molarity, which is the molar concentration of a solution, can be calculated by dividing the number of moles (n) by the volume (V).

That is;

Molarity (M) = n/V

According to the information provided in this question;

M = 2.40M

V = 56.0 mL = 56/1000 = 0.056 L

Since molarity = n/V

number of moles = M × V

n = 0.056 × 24

n = 1.34 mol

Answer: 1 mole =  6.02 x 10^ 23

Explanation:

The number of atoms in one mole of helium is equal to 6.022×10²³ He atoms.

Avogadro’s constant can be demonstrated as the proportionality factor that utilizes to count the number of particles such as ions, molecules, atoms, or ions in a sample of the substance.

Avogadro's number can be defined as the approximate count of nucleons in 1 gram of substance. The value of the Avogadro number is the mass of 1 mole of a compound, in grams, and is the number of nucleons in one particle.

The value of Avogadro’s number is equal to 6.022×10²³ per mole.

Given, the number of moles of the helium = 1 mol

Helium is a monatomic gas so it contains only one atom in its one molecule.

The number of atoms of Helium in one mole = 6.022 × 10²³

Therefore, the total number of atoms contained in a 1 mole sample of

helium is 6.022 × 10²³.

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Answer:

So confusing but I'll try

Answer:

No,  22.09%  is not a valid measurement

Explanation:

Precision has to do with how close a given set of measured values are to each other. It is quite different from accuracy. Accuracy refers to how close a given set of values is to the true value. A given set of values may be precise but not accurate and vice versa.

If we look at the values obtained;  22.09%,  22.15%, 22.18%, 22.23%, 22.25%, the value 22.09% is too far off the other values. This implies that it does not represent a valid measurement since it is not close to all the other values obtained.

Answer:

Explanation:

The equation we use to calculate the volume needed to prepare other [tex](C_1,V_1)[/tex] the solution that has a concentration [tex]C_2[/tex]  and volume [tex]V_2[/tex]  is:

[tex]C_1V_1 =C_2V_2[/tex]

[tex]V_1=\dfrac{C_2V_2}{C_1}[/tex]

where;

[tex]C_1[/tex]= concentration of the first solution

[tex]V_1[/tex] = volume of the first solution

[tex]C_2[/tex] = concentration of the second solution

[tex]V_2[/tex] = volume of the second solution

2) Reduction half cell reaction for the copper (II) ion is:

[tex]Cu^{2+} + 2e^- \to Cu[/tex]

3) [tex]Cu^{+2} + 2e^- \to Cu \text{ \ \ \ E = 0.3370}[/tex]

[tex]Zn^{+2} + 2 e^- \to Zn \ \ \ \ \ \ E = -0.763[/tex]

[tex]Zn \to Zn^{+2} + 2 e^- \ \ \ \ \ \ E = +0.763[/tex]

Since the reduction potential of Cu is more; it means copper will go into reduction and zinc will undergo oxidation.

Standard Potential =[tex]E^0_{left} - E^0_{right}[/tex]

[tex]= -0.763 -0.337[/tex]  ( since both are reduction potential)

[tex]\mathbf{E^0_{cell} = -1.100 volt}[/tex]

Cyclohexene is a cyclic, six-membered hydrocarbon that contains one double bond. The types of reactions that can occur in cyclohexene would be those that are typical with alkenes generally.

The pi-bonded electrons in the double bond are nucleophilic. So, electrophilic addition reactions could occur with cyclohexene. For example,

cyclohexene + HBr → bromocyclohexane

cyclohexene + H2O/H+ → cyclohexanol

cyclohexene + Br2 → trans-1,2-dibromocyclohexane (racemic)

The latter is a common test for alkenes where one adds bromine to a sample to see if there is decolorization, which would indicate the presence of nucleophilic pi bonds. Bromine, which is dark reddish-brown, will become clear as it reacts with an alkene to form a colorless haloalkane.

Cyclohexene can also be converted to the fully saturated cyclohexane by hydrogenation: cyclohexene + H2/Pd → cyclohexane.

Answer: The Spinal Cord

Explanation: The spinal cord runs along the dorsal side of the body and links the brain to the rest of the body.

Answer:

2. m = 426.6 gr

4. m = 143 gr

Answer:

5.56 x 10^23

Explanation:

Just convert and cancel out.

85 g N2O4 x 1 mol/92.01 g x 6.02 x 10^23 molecules /1 mol

Answer:Scientists have identified more than 260,000 kinds of plants. They classify plants according to whether they have body parts such as seeds, tubes, roots, stems, and leaves. The three main groups of plants are seed plants, ferns, and mosses.

Explanation:

hope i help

Answer:

[tex]ZnCO_3(s)+2KNO_3(aq)[/tex]

Explanation:

Hello there!

In this case, for the referred reactants side, we can evidence that this is a double displacement reaction whereby Zn gets bonded with CO3 and therefore K gets bonded with NO3 to yield:

[tex]Zn(NO_3)_2+K_2CO_3\rightarrow ZnCO_3+2KNO_3[/tex]

Whereas the KNO3 product is aqueous and the ZnCO3 is solid due to the corresponding solubility rules. Thereby, the answer is the third one.

Regards!

Answer:1 2 and 4

Explanation:

Answer:

C

Explanation:

It is run by Elon Musk, who is determined to get to Mars.

Answer:

C

Explanation:

because then you'd be able to make a group to raid, run with you, and save more people.

Answer:

5.86mol 2Fe(OH)3 * [3 mols Na SO4 / 2 mols Fe(OH)3] = 8.79 mols sodium sulfate

Explanation:

To solve this question you need to look at the ratio of sodium sulfate to iron(III) hydroxide. The ratio is 3 to 2, meaning that for every 3 moles of Na SO4 there are also 2 moles of Fe(OH)3. So, you can multiply 5.86 by (3/2) to find the mols of sodium sulfate.

1. The standard cell potential for this reaction is 0.14 V.

2. The standard cell potential for this reaction is 0.46 V.

3. The reduction potential for Hg2+(aq) + 2e^- → Hg(l) is 0.79 V.

1. The half-reactions are:

Oxidation: Sn2+(aq) → Sn(s) + 2e^-

Reduction: Pt2+(aq) + 2e^- → Pt(s)

To balance the charges, we multiply the oxidation half-reaction by 2:

2Sn2+(aq) → 2Sn(s) + 4e^-

Now, we can combine the half-reactions to form the overall cell reaction:

2Sn2+(aq) + Pt2+(aq) → 2Sn(s) + Pt(s)

The cell notation for this reaction is:

Sn(s) | Sn2+(aq) || Pt2+(aq) | Pt(s)

To calculate the standard cell potential (E°), we can look up the reduction potentials for each half-reaction. The reduction potential for Pt2+(aq) + 2e^- → Pt(s) is typically listed as 0.00 V. The reduction potential for Sn2+(aq) + 2e^- → Sn(s) is -0.14 V. The standard cell potential is the sum of the reduction potentials:

E° = E°(reduction) - E°(oxidation)

E° = 0.00 V - (-0.14 V) = 0.14 V

2. The half-reactions are:

Oxidation: Co2+(aq) → Co(s) + 2e^-

Reduction: Cr3+(aq) + 3e^- → Cr(s)

To balance the charges, we multiply the reduction half-reaction by 2:

2Cr3+(aq) + 6e^- → 2Cr(s)

Now, we can combine the half-reactions to form the overall cell reaction:

Co2+(aq) + 2Cr3+(aq) + 6e^- → Co(s) + 2Cr(s)

The cell notation for this reaction is:

Co(s) | Co2+(aq) || Cr3+(aq) | Cr(s)

To calculate the standard cell potential (E°), we look up the reduction potentials for each half-reaction. The reduction potential for Co2+(aq) + 2e^- → Co(s) is typically listed as -0.28 V. The reduction potential for Cr3+(aq) + 3e^- → Cr(s) is -0.74 V. The standard cell potential is the sum of the reduction potentials:

E° = E°(reduction) - E°(oxidation)

E° = -0.28 V - (-0.74 V) = 0.46 V

3. The half-reactions are:

Oxidation: Cr2+(aq) → Cr(s) + 2e^-

Reduction: Hg2+(aq) + 2e^- → Hg(l)

The balanced overall cell reaction is:

Cr2+(aq) + 2Hg2+(aq) + 4e^- → Cr(s) + 2Hg(l)

The cell notation for this reaction is:

Hg(l) | Hg2+(aq) || Cr2+(aq) | Cr(s)

To calculate the standard cell potential (E°), we look up the reduction potentials for each half-reaction. The reduction potential for Cr2+(aq) + 2e^- → Cr(s) is typically listed as -0.91 V.

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